IP=153 . 80 . 109 . 141 / 24


Mask =

Find the Answers for each

  1. What is the Dotted Decimal Subnet Mask?
  2. Find this SubNetwork
  3. Find the First useable IP address
  4. Find the Broadcast address
  5. Find the Last useable IP address
  6. What is the Total Number of IPs per Subnet?
  7. How many useable Hosts Addresses per subnet?
  8. What is the Next Subnetwork Number?
  9. What IP Class is this?
  10. How many possible Subnetworks with this IP Class and mask?

When completed Scroll down for the Answers

Tue, 30 Dec 2025 23:37:24 +0000
© by David Clauss Oct 2012















Repeat
Given IP153 . 80 . 109 . 141 / 24
#1. & 2.Find the SubNetwork
  1. Binary IP= 10011001 . 01010000 . 01101101 . 10001101 .
  2. With a 24 bit mask
  3. Dotted Decimal Subnet Mask = 255 . 255 . 255 . 0
  4. Binary Subnet Mask = 11111111 . 11111111 . 11111111 . 00000000

    Subnetwork Address = 153 . 80 . 109 . 0

#3.Find the First useable IP address 153 . 80 . 109 . 1
#4.Find the Broadcast address153 . 80 . 109 . 255
#5Find the Last useable IP address153 . 80 . 109 . 254
#6.What is the Total Number of IPs per Subnet?256
#7.How many useable Hosts address per subnet?254
#8.What is the Next Subnet Number?153 . 80 . 110 . 0
#9.What IP Class is this?

Class B

#10.How many possible Subnetworks with this class?
Bits this Class = 16
Bits Subnet Mask= 24

2^ 8 = 256

Also try: Networks Practice

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Tue, 30 Dec 2025 23:37:24 +0000